Difference between revisions of "2018 AMC 10A Problems/Problem 23"
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Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square <math>S</math> so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from <math>S</math> to the hypotenuse is 2 units. What fraction of the field is planted? | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square <math>S</math> so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from <math>S</math> to the hypotenuse is 2 units. What fraction of the field is planted? | ||
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Let the square have side length <math>s</math>. If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the big similar triangle is <math>\frac{5}{3}(2)=\frac{10}{3}</math>. Now, let's extend this big similar right triangle to the left until it hits the side of length 3. Now, the length is <math>\frac{10}{3}+s</math>, and using the ratios of the side lengths, the height is <math>\frac{3}{4}(\frac{10}{3}+s)=\frac{5}{2}+\frac{3s}{4}</math>. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so <cmath>\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } (\frac{2}{7})^2=\frac{4}{49}</cmath> | Let the square have side length <math>s</math>. If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the big similar triangle is <math>\frac{5}{3}(2)=\frac{10}{3}</math>. Now, let's extend this big similar right triangle to the left until it hits the side of length 3. Now, the length is <math>\frac{10}{3}+s</math>, and using the ratios of the side lengths, the height is <math>\frac{3}{4}(\frac{10}{3}+s)=\frac{5}{2}+\frac{3s}{4}</math>. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so <cmath>\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } (\frac{2}{7})^2=\frac{4}{49}</cmath> | ||
− | Now comes the easy part: finding the ratio of the areas: <math>\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\frac{145}{147}}</math> | + | Now comes the easy part: finding the ratio of the areas: <math>\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\frac{145}{147}.}</math> |
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+ | ==See Also== | ||
+ | {{AMC10 box|year=2018|ab=A|num-b=22|num-a=24}} | ||
+ | {{AMC12 box|year=2018|ab=A|num-b=16|num-a=18}} | ||
+ | {{MAA Notice}} |
Revision as of 18:35, 8 February 2018
Contents
Problem
Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from to the hypotenuse is 2 units. What fraction of the field is planted?
Solution
Let the square have side length . Connect the upper-right vertex of square with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is .
Square has area , and the two thin triangle regions have area and . The final triangular region with the hypotenuse as its base and height has area . Thus, we have
Solving gives . The area of is and the desired ratio is .
Solution 2
Let the square have side length . If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the big similar triangle is . Now, let's extend this big similar right triangle to the left until it hits the side of length 3. Now, the length is , and using the ratios of the side lengths, the height is . Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so
Now comes the easy part: finding the ratio of the areas:
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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